te')); return $arr; } /* 遍历用户所有主题 * @param $uid 用户ID * @param int $page 页数 * @param int $pagesize 每页记录条数 * @param bool $desc 排序方式 TRUE降序 FALSE升序 * @param string $key 返回的数组用那一列的值作为 key * @param array $col 查询哪些列 */ function thread_tid_find_by_uid($uid, $page = 1, $pagesize = 1000, $desc = TRUE, $key = 'tid', $col = array()) { if (empty($uid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('uid' => $uid), array('tid' => $orderby), $page, $pagesize, $key, $col); return $arr; } // 遍历栏目下tid 支持数组 $fid = array(1,2,3) function thread_tid_find_by_fid($fid, $page = 1, $pagesize = 1000, $desc = TRUE) { if (empty($fid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('fid' => $fid), array('tid' => $orderby), $page, $pagesize, 'tid', array('tid', 'verify_date')); return $arr; } function thread_tid_delete($tid) { if (empty($tid)) return FALSE; $r = thread_tid__delete(array('tid' => $tid)); return $r; } function thread_tid_count() { $n = thread_tid__count(); return $n; } // 统计用户主题数 大数量下严谨使用非主键统计 function thread_uid_count($uid) { $n = thread_tid__count(array('uid' => $uid)); return $n; } // 统计栏目主题数 大数量下严谨使用非主键统计 function thread_fid_count($fid) { $n = thread_tid__count(array('fid' => $fid)); return $n; } ?>javascript - Nodejs - Correct way to iterate through nested JSON array - Stack Overflow
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javascript - Nodejs - Correct way to iterate through nested JSON array - Stack Overflow

programmeradmin4浏览0评论

Consider the following sample JSON array:

[{
    info: {
        refOne: 'refOne',
        refTwo: [{
            refOne: 'refOne',
            refTwo: 'refTwo'
        }]
    }
}, {
    info: {
        refOne: 'refOne',
        refTwo: [{
            refOne: 'refOne',
            refTwo: 'refTwo'
        }]
    }
}]

The above JSON is a simple representation of a database query response, What is the correct way within Nodejs to loop through each 'refTwo' array within the parent info array?

sudo example: for each item in sample JSON for each refTwo item in current item do something

I have a suspicion that the 'async' lib may be required here but some advice is much appreciated.

Consider the following sample JSON array:

[{
    info: {
        refOne: 'refOne',
        refTwo: [{
            refOne: 'refOne',
            refTwo: 'refTwo'
        }]
    }
}, {
    info: {
        refOne: 'refOne',
        refTwo: [{
            refOne: 'refOne',
            refTwo: 'refTwo'
        }]
    }
}]

The above JSON is a simple representation of a database query response, What is the correct way within Nodejs to loop through each 'refTwo' array within the parent info array?

sudo example: for each item in sample JSON for each refTwo item in current item do something

I have a suspicion that the 'async' lib may be required here but some advice is much appreciated.

Share Improve this question edited Mar 28, 2014 at 10:18 Gabriel Llamas 18.4k26 gold badges90 silver badges117 bronze badges asked Mar 28, 2014 at 9:54 fearhsonicfearhsonic 931 gold badge2 silver badges6 bronze badges
Add a ment  | 

2 Answers 2

Reset to default 10

This is a simple javascript question:

var o = [...];

var fn = function (e){
    e.refOne...
    e.refTwo...
};

o.forEach (function (e){
    e.info.refTwo.forEach (fn);
});

You could use underscore or lodash to do it in a functional way.

For example have a look at Collections.each and Collections.map:

var _ = require('underscore');

var result = // your json blob here

var myRefs = _.map(results, function(value, key) {
  return value.info.refTwo;
};
// myRefs contains the two arrays from results[0].info.refTwo from results[1].info.refTwo now

// Or with each:
_.each(results, function(value, key) {
  console.log(value.info.refTwo);
}

// Naturally you can nest, too:
_.each(results, function(value, key) {
  _.each(value.info.refTwo, function(innerValue) { // the key parameter is optional
    console.log(value);
  }
}

Edit: You can of course use the forEach method suggested by Gabriel Llamas, however I'd remend having a look at underscore nonetheless.

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